Let
$$
W=2\Omega+(\nabla\times\mathbf u)\mathbin\cdot\mathbf e_z
$$
be the vertical <absolute vorticity>. Taking the vertical curl of the momentum equation removes both the tidal potential and the isothermal pressure force, because each is a <gradient field>. The two-dimensional <vorticity equation> is
$$
D_tW=-W\nabla\mathbin\cdot\mathbf u.
$$
The surface-density equation is
$$
D_t\Sigma=-\Sigma\nabla\mathbin\cdot\mathbf u.
$$
The <quotient rule> now gives
$$
D_t\left(\frac W\Sigma\right)
=\frac{D_tW}{\Sigma}-\frac W{\Sigma^2}D_t\Sigma=0.
$$
Since $\zeta=W/\Sigma$ is the <vortensity>,
$$
\boxed{D_t\zeta=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-321/3/a/solution